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Proof step 14 from proof sgn-one-01 (previous step)

sgn(true ' + bits0) = true ', false ∈ bits0 ⊦ bits0 = [], sgn(bits0) = true ', bits0.first

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits0, boolvar, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: sgn-base-01: ⊦ sgn(true ' + false ' + bits) = true ' + false ' + bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits0, boolvar, bits